Laplace transform of $f = t^2 -8t + 23$ for $t\geq 4$ and $0$ elsewhere $f$ is defined piecewise like so:
\begin{cases} 
      0 & t < 4 \\
      t^2 -8t + 23 & t \geq 4  \\
   \end{cases}
So using the stepwise function notation, $u_4(t) = u(t-4)$ (as in, the stepwise function shifted to the right by $4$), I find that $f = u_4(t) \cdot (t^2 -8t + 23) $
Now I want to use the formula for Laplace transforms of functions multiplied by stepwise functions:
$$\mathcal{L} (u_a(t) \cdot f(t)) = e^{-as} \mathcal L (f(t+a))$$
First I find $\mathcal L(f(t+a)):$
$\mathcal L f(t+5)=\mathcal L ((t+5)^2-8(t+5)+23) = \mathcal L(t^2+2t-8) = \frac2{s^3}  + \frac2{s^2} - \frac8s$
So  the final answer should be $e^{-4s} ( \frac2{s^3}  + \frac2{s^2} - \frac8s)$...but apparently this is wrong. Can anyone see why?
 A: Your answer is incorrect because you wrote $f(t+a)=f(t+5)$ when the step function had a right shift of $4$. Applying
$$\mathcal{L} \left(u(t-a)f(t)\right) = e^{-as} \mathcal {L} [f(t+a)],$$
we find
$$\mathcal{L}[f(t+4)]=\mathcal L \left[(t+4)^2-8(t+4)+23\right] = \mathcal {L}[t^2+7] = \frac2{s^3}  + \frac{7}{s}.$$
This gives the final answer of
$$\mathcal{L} \left(u(t-4)f(t)\right)=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$

I took a slightly different approach. We are given:
$$f(t)=\begin{cases} 
      0, & t < 4, \\
      t^2 -8t + 23, & t \geq 4,  \\
   \end{cases}=u(t-4)(t^2-8t+23).$$
Lets first complete the square
$$t^2-8t+23=(t-4)^2+7.$$
Then the problem becomes
$$f(t)=u(t-4)(t-4)^2+7u(t-4).$$
Applying
$$\mathcal{L}(t^2)=\frac{2}{s^{3}},\quad \mathcal{L}\big[u(t-c)\big]=\frac{e^{-cs}}{s},\quad \mathcal{L}\big[u(t-c)g(t-c)\big]=e^{-cs}\mathcal{L}[g(t)],$$
we find
$$\mathcal{L}[f(t)]=\mathcal{L}\left[u(t-4)(t-4)^2\right]+\mathcal{L}[7u(t-4)]=\frac{2e^{-4s}}{s^3}+\frac{7e^{-4s}}{s}=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$
A: Another way to do, only with the formal definition of the Laplace transform. Then 
$\mathcal{L}\lbrace f\rbrace=\displaystyle\int_{0}^{4} 0dt+\int_{4}^{\infty}e^{-st}(t^{2}-8t+23)dt$.
